We are given the following system of linear equations:

1. \( 3x + 4y = -3 \)
2. \( x + 2y = -1 \)

Let's solve this system using the substitution or elimination method. I'll use substitution for this problem.

### Step 1: Solve the second equation for \( x \)
From the second equation:
\[
x + 2y = -1
\]
Solve for \( x \):
\[
x = -1 - 2y
\]

### Step 2: Substitute \( x = -1 - 2y \) into the first equation
Substitute this expression for \( x \) into the first equation \( 3x + 4y = -3 \):
\[
3(-1 - 2y) + 4y = -3
\]

### Step 3: Simplify and solve for \( y \)
Distribute the \( 3 \) on the left-hand side:
\[
-3 - 6y + 4y = -3
\]
Combine like terms:
\[
-3 - 2y = -3
\]
Add 3 to both sides:
\[
-2y = 0
\]
Solve for \( y \):
\[
y = 0
\]

### Step 4: Substitute \( y = 0 \) back into the expression for \( x \)
Now, substitute \( y = 0 \) into \( x = -1 - 2y \):
\[
x = -1 - 2(0) = -1
\]

### Final Solution:
The solution to the system is:
\[
x = -1, \quad y = 0
\]

Would you like any further details or have any questions?

Here are five related questions:
1. What is the graphical interpretation of the solution to this system of equations?
2. How can we use the elimination method to solve this system instead?
3. What happens if the system of equations has no solution?
4. How can we check if the solution we found is correct?
5. Can we represent this system of equations in matrix form?

**Tip:** Always substitute the values of \(x\) and \(y\) back into the original equations to verify the solution!

This page offers a step-by-step guide on solving the linear system of equations 3x + 4y = -3 and x + 2y = -1 using the substitution method. The problem emphasizes key algebraic concepts and is suitable for students in Grades 9-11.

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